The remainder is 1. Since 7≡1(mod6)7\equiv 1\pmod{6}7≡1(mod6), every power 720267^{2026}72026 is also congruent to 1(mod6)1\pmod{6}1(mod6).

If you want the quick modular-arithmetic shortcut, it is:

  • 7≡1(mod6)7\equiv 1\pmod{6}7≡1(mod6)
  • 72026≡12026≡1(mod6)7^{2026}\equiv 1^{2026}\equiv 1\pmod{6}72026≡12026≡1(mod6)

So the answer is 1.